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pmt1_solution - IEOR 160 Homework#5 Solution 1 A concave...

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IEOR 160, Homework #5 Solution 1. A concave function cannot have a minimizer over an equality constrained feasible region. False. Consider function f ( x ) = ln x , over the feasible region with equality constraint x = 1. Then x = 1 is a minimizer. If the objective function of an optimization problem is convex and the feasible region is convex, then it is a minimization problem. False. Whether a problem is a maximization problem or minimization problem is nothing related to the property of objective function and feasible region. If f is continuously differentiable function, then all its local maxima is among its stationary points. True. If x is a local maximum of a concave function, then there exists a direction vector d for which the directional derivative at x is negative. False. f ( x ) = 1 is a concave function. And point x = 1 is a local maximum. But for any direction vector d , the directional derivative is 0. For a KKT point, if Lagrangian multiplier of a constraint is zero, then the constraint is inactive at this point. False. For a KKT point, Lagrangian multiplier is zero, the constraint can be either binding or not. For example, max x 2 , constraint is x 0. Then the KKT condition is 2 x + λ = 0, λx = 0, λ 0, x 0. Clearly, ( x, λ ) = (0 , 0) is a KKT point. And λ = 0, but the constraint x 0 is still active at this point. 2. Proof. Use contradiction. Assume that the statement is not true. That it, there is a local minima x of function f ( x ) on S and x is not a global minimum.
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